0
$\begingroup$

Recently, I am learning generalized inverse of a matrix. Given a real symmetric matrix $A\in\mathbb{R}^{n\times n}$ with ${\rm rank}(A)=r$, suppose the rank decomposition of $A$ is given as follows: $$ PAQ = \begin{bmatrix} I_{r} & 0\\ 0 & 0 \end{bmatrix}_{n\times n}, $$ where $P,Q\in\mathbb{R}^{n\times n}$ are invertible.

Through several examples, I observed the product $QP$ is always symmetric. Here comes my question: for any real symmetric matrix, is $QP$ also symmetric? Although I have tried a lot, I still can not give a rigorous proof. Any answer will be appreciated. Thanks a lot.

$\endgroup$
2
  • $\begingroup$ $QP$ must be symmetric in the specific case that $r = n$. I suspect that it must also be symmetric in the case that $r = n-1$. $\endgroup$
    – Ben Grossmann
    Commented Jun 13 at 14:56
  • $\begingroup$ Yes, if $r=n$, then $QP$ equals the inverse of $B$, which is also symmetric. $\endgroup$
    – W.J
    Commented Jun 13 at 15:10

1 Answer 1

Reset to default
1
$\begingroup$

Neither $PQ$ nor $QP$ is necessarily symmetric. As a counterexample, consider $$ A = \pmatrix{1&0&0\\0&0&0\\0&0&0}, \quad P = Q = \pmatrix{ 1&0&0\\ 0&1&1\\0&0&1}. $$ We find that $$ QP = PQ = P^2 = \pmatrix{1&0&0\\0&1&2\\0&0&1}, $$ which is not symmetric.

$\endgroup$
3
  • $\begingroup$ Thanks a lot for your quick answer. I find I have not prepared plenty of examples. $\endgroup$
    – W.J
    Commented Jun 13 at 15:08
  • $\begingroup$ @W.J Generating good examples is a skill you pick up. A common incorrect approach for a situation like this is to have a computer generate symmetric matrices at random, e.g. by selecting i.i.d. normally distributed entries. This is a bad idea here since matrices generated like this are invertible with nearly 100% probability. The key, however you're generating examples, is to look for a property to vary that you think might be relevant. In this case, making sure that you try symmetric matrices of different ranks is key. $\endgroup$
    – Ben Grossmann
    Commented Jun 13 at 15:15
  • $\begingroup$ Thanks again for your detailed instructions. I have tried rank-deficient matrix in math.stackexchange.com/q/4930822/692201 . In fact, I generated $A$ by $C^{T}$diag(1,...-1...,0,...)$C$, where $C$ is chosen to be invertible. Your example is succint and efficient. I forgot to choose $C$ as identity matrix. You help me a lot! $\endgroup$
    – W.J
    Commented Jun 13 at 17:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .